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Let A be an $n\times n$ matrix. Show that $\det(A^s) = (\det(A))^s$ for every $s\in \mathbb N = \{1,2,3,\cdots\}$

($\det$ is the determinant of the matrix).

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closed as off-topic by user223391, Dietrich Burde, Claude Leibovici, user91500, Jack's wasted life Nov 15 '16 at 6:56

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    $\begingroup$ Where are you stuck? Do you know that $\det(AB)=\det(A)\det(B)$? $\endgroup$ – carmichael561 Nov 14 '16 at 19:25
  • $\begingroup$ @carmichael561 yes I am aware of this but I'm unsure of how you can use this in this question. $\endgroup$ – Craig Nov 14 '16 at 19:26
  • $\begingroup$ Do you know how to use induction? $\endgroup$ – Paul Nov 14 '16 at 19:26
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    $\begingroup$ Taking $B=A$, you get $\det(A^2)=\det(A)^2$, for instance. In general, you can use induction. $\endgroup$ – carmichael561 Nov 14 '16 at 19:27
  • $\begingroup$ So would you then do det(A²B) = det(A^3) = det(A)^3 and so on? $\endgroup$ – Craig Nov 14 '16 at 19:30
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Proceed by induction: For $s=1$, it's trivial. Assume true for $s=k$, prove for $s=k+1$. Then

$\det{A^{k+1}}= \det{(A^k)} det A=(\det{A})^k\det{A}=(\det{A})^{k+1}$

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